a) $\sqrt{4x + 8}$ + 4$\sqrt{\frac{x + 2}{4}}$ - $\sqrt{9x + 18}$ = 0 Đkxđ: x ≥ -2
⇔ 2$\sqrt{x + 2}$ + 2$\sqrt{x + 2}$ - 3$\sqrt{x + 2}$ = 0
⇔ $\sqrt{x + 2}$ = 0
⇔ x + 2 = 0
⇔ x = -2 (t/m)
Vậy x = -2
b) $\sqrt{x² - x - 6}$ = $\sqrt{x - 3}$ Đkxđ: x ≥ 3
⇔ x² - x - 6 = x - 3
⇔ x² - 2x - 3 = 0
⇔ x² - 3x + x - 3 = 0
⇔ x.(x - 3) + (x - 3) = 0
⇔ (x - 3)(x + 1) = 0
⇔ \(\left[ \begin{array}{l}x - 3 = 0\\x + 1 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 3 (T/m)\\x = -1 (Không t/m)\end{array} \right.\)
Vậy x = 3
c) $\sqrt{\frac{x - 7}{x - 3}}$ = $\sqrt{x + 3}$ Đkxđ: -3 < x < 3; x ≥ 7
⇔ $\frac{x - 7}{x - 3}$ = x + 3
⇔ x - 7 = (x - 3)(x + 3)
⇔ x - 7 = x² - 9
⇔ x² - x - 2 = 0
⇔ x² - 2x + x - 2 = 0
⇔ x.(x - 2) + (x - 2) = 0
⇔ (x - 2)(x + 1) = 0
⇔ \(\left[ \begin{array}{l}x - 2 = 0\\x + 1 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 2 (t/m)\\x = -1 (t/m)\end{array} \right.\)
Vậy x ∈ {2; -1}
